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%\author{王立庆（2019级数学与应用数学1班）}
\author{学号 \underline{\hspace{4cm}} 姓名  \underline{\hspace{4cm}} }
%\title{高等代数第六章：向量空间}
\title{第八章欧氏空间（8.3-8.4）考试解答 }
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\date{2023年5月17日}

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\begin{document}

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%\begin{abstract}
%%主要内容：
%7.3. 
%7.4. 
%7.5. 

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\begin{enumerate}

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\item %1
设 $V=\mathbb{R}^2$ 是平面，设 $\sigma: V\to V$ 是关于直线 $y=2x$ 的反射。
\begin{enumerate}
\item  设 $\alpha=\begin{bmatrix} x\\ y \end{bmatrix}$, 写出 $\sigma(\alpha)$ 的具体表达式。
\item  证明 $\sigma$ 是线性变换，并写出 $\sigma$ 关于标准基 $\{\varepsilon_1, \varepsilon_2\}$ 的矩阵。
\item  验证 $\langle \sigma(\alpha), \sigma(\alpha) \rangle = \langle \alpha,\alpha \rangle$, 从而证明 $\sigma$ 是正交变换。
\end{enumerate}

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{\color{red}解答：
\begin{enumerate}

\item  
\begin{enumerate}
\item  设向量 $\alpha=\begin{bmatrix} x\\ y \end{bmatrix}$ 在直线 $y=2x$ 上的正射影为 $\alpha_1=\begin{bmatrix} t\\ 2t \end{bmatrix}$. 
\item  将条件 $(\alpha-\alpha_1) \perp \alpha_1$ 写成内积形式，即为 $\langle \alpha-\alpha_1, \alpha_1 \rangle =0$, 由此求得 $\alpha_1= \frac{1}{5}\begin{bmatrix} x+2y\\ 2x+4y \end{bmatrix}$. 
\item  因此 $\sigma(\alpha) = \alpha + 2(\alpha_1-\alpha) = 2\alpha_1-\alpha = \frac{1}{5} \begin{bmatrix} -3x+4y \\  4x+3y \end{bmatrix}$. 
\end{enumerate}

\item  
\begin{enumerate}
\item  因为 $\sigma \left( \begin{bmatrix} x\\ y \end{bmatrix} \right) 
= \begin{bmatrix} - \frac{3}{5} & \frac{4}{5} \\  \frac{4}{5} & \frac{3}{5} \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} $, 
即 $\sigma(\alpha) = A\alpha$, 其中 $A$ 是二阶矩阵，所以 $\sigma$ 是线性变换。
\item  因为 $(\sigma(\varepsilon_1), \sigma(\varepsilon_2))
= (\varepsilon_1,\varepsilon_2) \cdot \begin{bmatrix} - \frac{3}{5} & \frac{4}{5} \\  \frac{4}{5} & \frac{3}{5} \end{bmatrix}$, 
所以 $\sigma$ 关于标准基的矩阵为 
$A=\begin{bmatrix} - \frac{3}{5} & \frac{4}{5} \\  \frac{4}{5} & \frac{3}{5} \end{bmatrix}$. 
 \end{enumerate}
 
\item  
\begin{enumerate}
\item  计算可得 $\langle \sigma(\alpha), \sigma(\alpha) \rangle = \frac{1}{25} [(-3x+4y)^2+(4x+3y)^2] = x^2+y^2$. 
\item  另一方面 $\langle \alpha,\alpha \rangle = x^2+y^2$, 因此 $\langle \sigma(\alpha), \sigma(\alpha) \rangle = \langle \alpha,\alpha \rangle$. 
\item  根据定义，可得 $\sigma$ 是正交变换。
\end{enumerate}

\end{enumerate}

}

\vspace{0.2cm}

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\item %2
设 $V=\mathbb{R}^2$ 是平面，设 $\sigma: V\to V$ 由 
$\sigma \left( \begin{bmatrix} x\\ y \end{bmatrix} \right) 
= \begin{bmatrix} 2&2 \\ 2&5 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} $ 定义。
\begin{enumerate}
\item  按定义验证 $\sigma$ 是对称变换。
\item  将矩阵 $A=\begin{bmatrix} 2&2 \\ 2&5 \end{bmatrix}$ 正交相似于对角矩阵。
\item  求规范正交基 $\{\gamma_1, \gamma_2\}$ 使得 $\sigma$ 关于这个基的矩阵是对角矩阵。
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  设 $\alpha=\begin{bmatrix} x\\ y \end{bmatrix}$, $\beta=\begin{bmatrix} u\\ v \end{bmatrix}$,  计算可知 
\begin{eqnarray*}
\langle \sigma(\alpha),\beta \rangle &=& \langle \begin{bmatrix} 2x+2y \\ 2x+5y \end{bmatrix}, \begin{bmatrix} u\\ v \end{bmatrix} \rangle 
= (2x+2y)u + (2x+5y)v, \\ 
\langle \alpha,\sigma(\beta) \rangle &=& \langle \begin{bmatrix} x \\ y \end{bmatrix}, \begin{bmatrix} 2u+2v \\ 2u+5v \end{bmatrix} \rangle 
= x(2u+2v) + y(2u+5v). 
\end{eqnarray*}
可见 $\langle \sigma(\alpha),\beta \rangle = \langle \alpha,\sigma(\beta) \rangle$, 因此 $\sigma$ 是对称变换。

\item  
\begin{enumerate}
\item  先求特征多项式， $f(\lambda) = \det(\lambda E-A) = \begin{vmatrix} \lambda-2&-2 \\ -2&\lambda-5 \end{vmatrix} = \lambda^2-7\lambda+6$. 
\item  然后求特征值， 设 $f(\lambda)=0$, 求得 $\lambda_1=1$, $\lambda_2=6$. 
\item  再求特征向量，对 $\lambda_1=1$, 
求解 $\begin{bmatrix} -1&-2 \\ -2&-4 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$, 
可得 $\begin{bmatrix} x\\ y \end{bmatrix} = k\begin{bmatrix} 2\\ -1 \end{bmatrix}$, $k\in \mathbb{R}$.  \\ 
对 $\lambda_2=6$, 
求解 $\begin{bmatrix} 4&-2 \\ -2&1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$, 
可得 $\begin{bmatrix} x\\ y \end{bmatrix} = k\begin{bmatrix} 1\\ 2 \end{bmatrix}$, $k\in \mathbb{R}$. 
\item  属于每个特征值的线性无关的特征向量都是1个，所以略过特征向量的正交化步骤。
\item  将不同特征值的特征向量规范化，
取 $\gamma_1=\frac{1}{\sqrt{5}}\begin{bmatrix} 2\\ -1 \end{bmatrix}$, $\gamma_2=\frac{1}{\sqrt{5}}\begin{bmatrix} 1\\ 2 \end{bmatrix}$. 
\item  因此取正交矩阵 $U=\frac{1}{\sqrt{5}} \begin{bmatrix} 2&1 \\ -1&2 \end{bmatrix}$, 则有 
$U^tAU = \begin{bmatrix} 1&0 \\ 0&6 \end{bmatrix}$. 

\end{enumerate}

\item  
\begin{enumerate}
\item  所求规范正交基可取为 
$\{ \gamma_1=\frac{1}{\sqrt{5}}\begin{bmatrix} 2\\ -1 \end{bmatrix}, \gamma_2=\frac{1}{\sqrt{5}}\begin{bmatrix} 1\\ 2 \end{bmatrix}\}$. 
\item  其它符合条件的规范正交基为 $\{\gamma_1, -\gamma_2\}$, $\{-\gamma_1, \gamma_2\}$, $\{-\gamma_1, -\gamma_2\}$ 等。

\end{enumerate}

\end{enumerate}

}

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\end{enumerate}


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